LeetCode 26 – Python 3 (Week 5 – 05)

Solution 1

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        length = 0
        for i in range(len(nums)):
            if nums[length] != nums[i]:
                length += 1                
                nums[length] = nums[i]
                
        return length + 1

Time complexity is O(n). Space complexity is O(1).

Solution 2

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        i = 0
        while i < len(nums)-1:
            if nums[i] == nums[i+1]:
                nums.remove(nums[i+1])
            else: i += 1
                
        return len(nums)

Time complexity is O(n). Space complexity is O(1).

Leave a comment

Design a site like this with WordPress.com
Get started